If $\sqrt n X_n\xrightarrow{d}N(0,1)$, what can we say about $X_n$? 
Suppose that $X_1,X_2,\ldots$ are random variables such that
  $$
\sqrt nX_n\xrightarrow{d} N(0,1)\quad\text{as}\quad n\to\infty.
$$
  What can we say about the convergence of $X_n$ in some sense as $n\to\infty$?

Since $\sqrt n\to\infty$, $X_n$ has to go to $0$ in an appropriate sense, right? Can we deduce that $X_n$ goes to zero almost surely? Or does it only have to go to $0$ in probability? For the convergence in distribution, the random variables need not be defined on the same probability space, so I am not sure if almost sure convergence or convergence in probabiltiy is even appropriate in this situation.
Suppose that $Y,Y_1,Y_2,\ldots$ are iid random variables with $\operatorname EY=0$ and $\operatorname EY^2=1$. Set $X_n=n^{-1}\sum_{k=1}^nY_k$. Then $\sqrt nX_n\xrightarrow{d}N(0,1)$ and $X_n\to0$ almost surely. Hence, we have an example when $X_n\to0$ almost surely. However, I'm not sure if it is possible to construct an example when $X_n\to0$ only in probability.
Any help is much appreciated!
 A: Let $N$ be a standard normal random variable. Let $X_n := n^{-1/2} \max\{-n^{1/4}, \min\{n^{1/4}, N\}\}$. Then $(X_n)$ satisfies the hypotheses of the question, and converges to $0$ in $\mathbb{L}^\infty$ (so almost surely, in $\mathbb{L}^p$...). This is somewhat artificial, but it shows that you can get convergence in very strong ways, more than you can hope the partial sums of a i.i.d. sequence.
Let $(U_n)$ be a sequence of i.d.d. uniform random variables, independent from $N$. Let $Y_n := X_n$ if $U_n \geq 1/n$, and $Y_n = n$ otherwise. Then :
a) $\|Y_n\|_{\mathbb{L}^p} \geq n^{1-\frac{1}{p}}$, so $Y_n$ does not converge to $0$ in $\mathbb{L}^p$ for any $p \in [1, \infty]$.
b) By the Borel-Cantelli lemma, almost surely, $Y_n = n$ infinitely often, so $Y_n$ does not converge to $0$ almost surely.
c) $\mathbb{P} (Y_n \neq X_n)$ converges to $0$, so $(\sqrt{n}Y_n)$ has the same limit in distribution as $(\sqrt{n} X_n)$, that is, $N$.
To finish, I'll answer your remark :

For the convergence in distribution, the random variables need not be defined on the same probability space, so I am not sure if almost sure convergence or convergence in probabiltiy is even appropriate in this situation.

This is a good remark. To simplify, when speaking about sequence of random variables, we often assume that all random variables are defined on the same probabilized space. However, convergence in distribution makes sense even if it is not the case, and indeed is sometimes used that way. However, we do need to keep the same measured space to speak about almost sure convergence.
A first answer is that, as a result, we often cannot upgrade a result about convergence in distribution to a result about almost sure convergence. The simplest example would be, for instance, the difference between $X_n := N$ (choose $N$, then the sequence $(X_n)$ is constant) and $(X_n)$ i.i.d. with the same distribution as $N$. The trajectories, and thus the convergence properties, are very different for the first sequence and for the second, but the distribution of each $X_n$ is always the same.
A second answer is that sometimes, we can say that a property holds for any realization of $(X_n)$ on a single probability space. For instance, if you take $A_n$ an event which depends only on $X_n$, and with $\sum_n \mathbb{P} (X_n \in A_n) < +\infty$, then, by the Borel-Cantelli lemma, no matter how you realize the random variables $X_n$ (they may be independent or not), almost surely, only finitely many $A_n$ shall occur. This shows that the almost sure behavior is sometimes constrained by the distributions of the marginals ; that said, it would still be improper to talk about almost sure convergence if the $(X_n)$ are not defined on the same probabilized space.
